# Acceleration under gravity

1. Apr 29, 2014

### Govind_Balaji

1. The problem statement, all variables and given/known data
A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take to reach the ground?

Ans:4.3 seconds
2. Relevant equations

3. The attempt at a solution

I can't understand the question.
Is the initial velocity of the ball when it is dropped=0? or is it 7m/s.

What value should I take for g? Should I take g= 9.8 m/s$^2$?
Please help me. I assumed ball is dropped from 60 m height with initial velocity=0 and a=g=9.8 m/s$^2$. But it gives wrong answer. What is the concept of this question? How to solve it?

2. Apr 29, 2014

### Staff: Mentor

g = 9.8 m/s2 is an acceptable value for g unless otherwise instructed.

The ball is being dropped (which means simply released) from a moving platform. It shares the motion of the platform until the instant it is released after which its trajectory is independent ("free fall").

3. Apr 29, 2014

### Govind_Balaji

Let v be the final velocity when it reaches the ground.
Let x be the distance=60 m
Let u be the initial velocity=0 m/s

a=g=9.8 m/s

By 2nd equation of motion,
$x=ut+\frac{at^2}{2}$
$x=0*t+\frac{9.8*t^2}{2}$
$60=\frac{9.8*t^2}{2}$
$120=9.8t^2$
$t^2=\frac{120}{9.8}$
$t^2=2.44$
$t=\sqrt{2.44}$
$t=1.56$

Actually I should get $t=4.3$

4. Apr 29, 2014

### Simon Bridge

What arguments from the problem statement can you make for either possibility? Something to do with the velocity of the balloon?

Also remember that velocity is a vector.

Acceleration is also a vector - is the acceleration of the ball in the same direction as the initial velocity?

If you make "downwards" positive - what are the values for initial velocity and acceleration going to be.

5. Apr 29, 2014

### Staff: Mentor

Reconsider the initial velocity of the ball.

6. Apr 29, 2014

### Govind_Balaji

I wrote everything that was in my text book. I said earlier that I didn't understand the question. I assumed the initial velocity to be 0. I am not certain.

7. Apr 29, 2014

### Govind_Balaji

Yes I made downwards positive since it was all about downwards motion.

8. Apr 29, 2014

### Govind_Balaji

My friend said that momentum of the balloon slows down the velocity of the ball. I think it is not right. Am I correct?

9. Apr 29, 2014

### Simon Bridge

I can see you are struggling to understand the problem - but if you do not explain how you are thinking about it, we can only guess what you are having trouble with. Don't worry about sounding silly - we've all done that: we understand.

OK - imagine someone in the balloon holding the ball out over the side.
The balloon is going upwards, before the ball is released: what is the ball doing?

At the instant the ball is released, it still has the same velocity as just before it was released.
Now do you see?

If "downwards" is positive, what is the sign of the initial velocity of the ball?

10. Apr 29, 2014

### Govind_Balaji

Thank you both, I found that initial velocity=-7 m/s. I also learnt that in a moving platform, an object in the platform shares the velocity with the platform. I found this by comparing me travelling in a train.

Here's my new attempt:

$$u=-7 m/s$$

$x=60 m$

$g=9.8 m/s^2$

$x=ut+\frac{at^2}{2}$
$60=-7t+\frac{(9.8)t^2}{2}$
$60=\frac{-14t+9.8t^2}{2}$
$120=-14t+9.8t^2$
Using quadratic equation formula, I got t=4.28$\approx$4.3 s

11. Apr 29, 2014

### Simon Bridge

Well done:
You'd have got +4.3s and -2.9s ... you want the time that is in the future.
When you do long answers you should show that step.

BTW: good LaTeX use.

Reality check:
This time should be longer than if the ball was just dropped from a stationary balloon.
The calculation neglects air resistance. IRL air resistance cannot be neglected for such a long fall.

12. Apr 30, 2014

### Govind_Balaji

Thank you. actually I started typing the answer before your last post. It took a long time to type LaTex(15-20 mins!!!)

13. Apr 30, 2014

### Simon Bridge

It gets faster with practice.

BTW the "itex" boxes are for inline use (where an equation sits inside a paragraph), while "tex" boxes are used for display equations - those that get their own line. The display form is best when you have lots of exponents or fractions.

i.e. a quadratic $ax^2+bx+c=0$ (1) can be solved by: $$x\in\frac{-b\pm\sqrt{b^2-4ac}}{2a}\qquad \small{\text{...(2)}}$$... (1) is the "standard form" and (2) is the "quadratic equation".

If I put (2) in the inline form, it comes out like $x\in\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

You can do multi-line equations... watch this:

\begin{align} & & ax+by=e & \qquad ...(1) \\ & & cx+dy=f & \qquad ...(2) \\ (1)\rightarrow & & y=\frac{e-ax}{b} & \qquad ...(3)\\ (3)\rightarrow (2) & & cx+d\frac{e-ax}{b}=f & \\ & & \implies x =\frac{bf-de}{bc-ad} & \qquad ...(4) \end{align}

Of course (1) and (2) can be written as a matrix equation:
$$\begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} x\\ y\end{pmatrix} =\begin{pmatrix} e\\ f\end{pmatrix}$$
... just showing off :)

Last edited: Apr 30, 2014